Other than bashing the problem with brute force (Mathematica doesn't take very long to try all $8!=40320$ possible permutations of the integers 1 through 8), I'd start by asking which pairs of distinct numbers from $\{1,2,3,4,5,6,7,8\}$ $a$ and $b$ satisfy $a-b=5$.
Further hint on that point:
If $a\le5$, then $b\not\in\{1,2,3,4,5,6,7,8\}$, ...
And where I'd go from there:
What about $c$ and $d$? How do the limitations on $a$ and $b$ interact with those on $c$ and $d$?
edit: Since a nearly-complete answer has been posted in another answer, here's the Mathematica code I used to solve it by brute-force:
#[[1]] & /@
Select[Table[{l,
l[[1]] - l[[2]] == 5 && l[[3]] - l[[4]] == 2 &&
l[[5]] - l[[6]] == 2 && l[[7]] - l[[8]] == 3}, {l,
Permutations[Range[8]]}], #[[2]] &]
and the solutions:
$a=7$, $b=2$, $c=3$, $d=1$, $e=6$, $f=4$, $g=8$, $h=5$
$a=7$, $b=2$, $c=5$, $d=3$, $e=8$, $f=6$, $g=4$, $h=1$
$a=7$, $b=2$, $c=6$, $d=4$, $e=3$, $f=1$, $g=8$, $h=5$
$a=7$, $b=2$, $c=8$, $d=6$, $e=5$, $f=3$, $g=4$, $h=1$
Also, here's how I'd solve it by hand:
The only possibilities for $a$ and $b$, based on the hints discussed above, are $(a,b)=(6,1)$, $(a,b)=(7,2)$, or $(a,b)=(8,3)$. Though I'd suggested looking at $c$ and $d$ next, I'd actually look at $g$ and $h$ next, since their difference ($3$) is the next largest difference after $a$ and $b$.
If $(a,b)=(6,1)$, then $(g,h)$ could be $(5,2)$, $(7,4)$, or $(8,5)$. On paper at this point, for these three cases, I'd have: $\not1\;\not2\;3\;4\;\not5\;\not6\;7\;8$; $\not1\;2\;3\;\not4\;5\;\not6\;\not7\;8$; and $\not1\;2\;3\;4\;\not5\;\not6\;7\;\not8$ (respectively). The remaining pairs of numbers ($c$ and $d$, $e$ and $f$) each differ by two and it should be clear from these lists with the already-used numbers crossed out that none of them permit two such pairs.
If $(a,b)=(7,2)$, then $(g,h)$ could be $(4,1)$, $(6,3)$, or $(8,5)$. On paper at this point, for these three cases, I'd have: $\not1\;\not2\;3\;\not4\;5\;6\;\not7\;8$; $1\;\not2\;\not3\;4\;5\;\not6\;\not7\;8$; and $1\;\not2\;3\;4\;\not5\;6\;\not7\;\not8$ (respectively). In the first case, $(c,d)$ and $(e,f)$ can be $(5,3)$ and $(8,6)$ (in either order); the second case does not allow two pairs of numbers with difference 2; in the third case, $(c,d)$ and $(e,f)$ can be $(3,1)$ and $(6,4)$ (in either order). This gives 4 solutions:
$a=7$, $b=2$, $c=3$, $d=1$, $e=6$, $f=4$, $g=8$, $h=5$
$a=7$, $b=2$, $c=6$, $d=4$, $e=3$, $f=1$, $g=8$, $h=5$
$a=7$, $b=2$, $c=5$, $d=3$, $e=8$, $f=6$, $g=4$, $h=1$
$a=7$, $b=2$, $c=8$, $d=6$, $e=5$, $f=3$, $g=4$, $h=1$
If $(a,b)=(8,3)$, then $(g,h)$ could be $(4,1)$, $(5,2)$, or $(7,4)$. On paper at this point, for these three cases, I'd have: $\not1\;2\;\not3\;\not4\;5\;6\;7\;\not8$; $1\;\not2\;\not3\;4\;\not5\;6\;7\;\not8$; and $1\;2\;\not3\;\not4\;5\;6\;\not7\;\not8$ (respectively). It should be clear from these lists with the already-used numbers crossed out that none of them permit two pairs with difference 2.